用Ajax调用JavaScript函数

Question

I'm using a multiselect dropdown to filter images stored in my db. The results are paginated, however the pagination fails once the results are filtered. The filter uss ajax to make a php call to the database.

What I think is happening is that the once the results are loaded in the div the paginate javascript function has already fired and wont a second time. Is there a way to call the function everytime the results are filtered?

I believe I just need to recall this each time:

    <script type="text/javascript">

jQuery(function($){

    $('ul#items').easyPaginate({
        step:6
    });

});    

    </script>

Ajax call:

<script>
function filterResults(sel)
{
    var selectedOptions = [];
    for (var i = 0; i < sel.options.length; i++) {
        if (sel.options[i].selected) {
            selectedOptions.push("'" + sel.options[i].value + "'");
        }
    }
    if (selectedOptions.length == 0) {
        document.getElementById("divResults").innerHTML="";
        return;
    }
    str = selectedOptions.join(",");
    if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    } else  {// code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange=function() {
        if (xmlhttp.readyState==4 && xmlhttp.status==200) {
            document.getElementById("divResults").innerHTML=xmlhttp.responseText;
        }
    }
    xmlhttp.open("GET","filter_ajax.php?filter="+str,true);
    xmlhttp.send();
}

</script>

ajax_filter.php:

<?php

    include ("connect.php");

    $filter = $_GET["filter"];
    $filterIn = $filter;

        $result = mysql_query("SELECT * FROM edt_images
                                WHERE category1 IN ($filterIn)
                                OR category2  IN ($filterIn)
                                OR category3  IN ($filterIn)
                                OR category4 IN ($filterIn)
                                OR category5 IN ($filterIn)
                                OR category6 IN ($filterIn)")                           
            or die(mysql_error());

        echo "<div id='results_container'>";
        echo "<ul id='items'>";

        while ($row = mysql_fetch_array($result)) {

            echo "<li><img src='files/300x200/thumb2_".$row['item_name'].".".$row['file_extension']."' class='filtered_images' border='0'/></li>"; 
        }
        echo "</ul>";
        echo "</div>";
?>

Answer 1

If you are using jQuery you can simplify the code within the filterResults function greatly by utilizing the framework it provides. I would read up a bit here as you will be amazed by the extensive functionality.

This code is the jQuery equivalent of your previous code,

   function filterResults(sel) {
        var selectedOptions = [];
        for (var i = 0; i < sel.options.length; i++) {
            if (sel.options[i].selected) {
                selectedOptions.push("'" + sel.options[i].value + "'");
            }
        }
        if (selectedOptions.length == 0) {
            $("divResults").html("");
            return;
        }
        filteStr = selectedOptions.join(",");
        $.ajax({
            url: "http://www.google.com",
            type: "get",
            dataType: "html",
            data: {
                filter: filteStr
            },
            success: function (responseHtml) {
                $("divResults").html(responseHtml);
                //You can put your code here
                $('ul#items').easyPaginate({
                    step: 6
                });
            },
            error: function (responseHtml) {
                //Handle error
            }
        });
    }

To answer the question the code within the success callback of the ajax call will be executed upon a receiving the data back from the server. This code above should work as expected.

Almost everything to know lies here and here . ;)

Answer 2

Well

According to the comments I'll have to study up on javascript and jquery, never really was to concerned with it before.

Anyways by recalling the javascript function

if (xmlhttp.readyState==4 && xmlhttp.status==200) {
            document.getElementById("divResults").innerHTML=xmlhttp.responseText;
            RESTATING HERE
}

I was able to get it working.

Answer 3

Put javascript call into your php-ajax echo, example:

<?
$msg = "hello";
?>
<script type="text/javascript">alert("<?=$msg?>"></script>